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# Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
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(s+1/2)2- 9/4
3 2
to be the same as that found above.
f(t) = (2/3)u2(t)e (t2)/2 sinh—(t-2), which can be shown
22
By completing the square in the denominator of F we can write F(s) = (2s+1)/[(2s+1) + 4]. This has the form
G(2s+1) where G(u) = u/(u +4). We must find ?-1{G(2s+1)}. Applying the results of Problem 19(c), we
have ?-1{F(s)} = —e-t/2 cos(-t).
22
If the approach of Problem 21 is used we find
f(t) = (1/3)e2t/3 sinh(t/3), which is equivalent to the given answer using the definition of sinht.
Assuming that term-by-term integration of the infinite series is permissible and recalling that ?{uc(t)} = e-cs/s
for s > 0, we have ?{f(t)} = (1/s) + (-1)k ?{uk(t)}
k=1
• •
= (1/s) + ^^(-1)k e-ks/s = ?(-0 k]/s. We recognize k=1 k=0
the last infinite series as the geometric series, ^ark,
k=0
with a = 1 and r = -e s. This series converges to [1/(1+e-s) ] if |r| < 1. Hence,
?{f(t)} = (1/s)[1/(1+e-s)], s > 0.
Section 6.4
111
28. Using the definition of the Laplace transform we have
F(s) = ?{f(t)} = J0 e-stf(t)dt. Since f is periodic with period T, we have f(t+T) = f(t). This suggests that we
f -st
rewrite the improper integral as I e f(t)dt =
^ Mn+1)T_st
y InT e f(t)dt. The periodicity of f also suggests
n=0
that we make the change of variable t = r + nT. Hence,
“ T “ T
F(s) = Jo e-s(r+nT)f(r+nT)dr = (e sT) n J0 e-rsf(r)dr,
n=0 n=0
where we have used the fact that
f(r+nT) = f(r+(n-1)T) = ... = f(r+T) = f(r) from the definition that f is periodic. We recognize this last

series as the geometric series, aun, with
n=0
T
-rs -sT
a = |0e f(r)dr and u = e . The geometric series
converges to a/(1-u) for |u| < 1 and consequently we obtain
-sT -1 T -rs
F(s) = (1 - e-sT)-1 |o e-rsf(r)dr, s > 0.
30. The function f is periodic with period 2. The result of
2 -st -2s
Problem 28 gives us ?{f(t)} = I e^fdddt/d-e- s). Calculating the integral we have
•2 i*1 i*2
J0 e stf(t)dt = J0 e stdt - J1 e stdt
2s -s
= d-e^/s + (e s-es)/s = (e^^e^+D/s
-s 2
= (1-e ) /s. Since the denominator of ?{f(t)}, 1 - e-2s, may be written as d-e^d+e-6) we obtain the desired answer.
Section 6.4, Page 321
1. f(t) can be written in the form f(t) = 1 - up/2(t) and thus the Laplace transforms of the D.E. is (s2+1)Y(s) - sy(0) - y'(0) = (1/s) - e-Ps/2/s. Introducing the I.C. and solving for Y(s), we obtain
Y(s) = (s2+1) -1 + [s(s2+1)] -1 - e-Ps/2/s(s2+1). Using partial fractions on the second and third terms we find
112
Section 6.4
Y(s) = (1/s) + (s2+1) -1 - s/(s2+1) - e-Ps/2/s + e-Ps/2s/(s2+1). The inverse transform of the first three terms can be obtained directly from Table 6.2.1. Using Theorem 6.3.1 to find the inverse transform of the last two terms we
have ?- 1{e-ps /2/s} = up/2(t)g(t - p/2) where g(t) = ?-1{1/s} = 1 and
?-1{e Ps/2s/(s2+1)} = up/2(t)h(t - p/2) where
h(t) = ?-1{s/(s2+1)} = cost. Hence, y = 1 + sint - cost + up/2(t)[cos(t - p/2) - 1]
= 1 + sint - cost - up/2(t)[1 - sint]. The graph of the forcing function is a unit pulse for 0 ? t < p/2 and 0 thereafter. The graph of the solution will be composed of two segments. The first, for 0 ? t < p/2, is a sinusoid oscillating about 1, which represents the system response to a unit forcing function and the given initial conditions. For t > p/2, the forcing function, f(t), is zero and the "initial" conditions are y(p/2) = lim 1 + sint - cost = 2 and
t—nr/2
y'(p/2) = lim cost + sint = 1. In this case the system
t—nr/2
response is y(t) = 2sint - cost, which is a sinusoid oscillating about zero.
3. According to Theorem 6.3.1,
?{u2p(t)sin(t-2p)} = e-2ps ?{sint} = e-2ps/(s2+1). Transforming the D.E., we have
(s2+4)Y(s) - sy(0) - y'(0) = 1/(s2+1) - e-2ps/(s2+1). Introducing the I.C. and solving for Y(s), we obtain -2 ps 2 2
Y(s) = (1-e s/(s +1)(s +4). We apply partial fractions to write
x i- 2 - -1 i 2 „ _1 -2ps 2 - -1 -2ps ^
Y(s) = [s +1) - (s +4) - e (s +1) + e (s +4) ]/3.
We compute the inverse transform of the first two terms directly from Table 6.2.1 after noting that
2 -1 2 (s +4) = (1/2)[2/(s +4)]. We apply Theorem 6.3.1 to the
last two terms to obtain the solution,
y =(1/3){sint-(1/2)sin2t-u2p(t)[sin(t-2p)-(1/2)sin2(t-2p)]}.
This may be simplified, using trigonometric identities, to y = [(2sint - sin2t)(1-u2p(t))]/6. Note that the forcing function is sint - sin(t-2p) = 0 for t > 2p. The solution is y(t) = 2sint - sin2t for 0 ? t < 2p. Thus
y(2p ) = 0 and y'(2p ) = 2cos2p - 2cos4p = 0. Hence the "initial" value problem for t > 2p is y" + 4y = 0, y(2p) = 0, y'(2p) = 0, which has the trivial solution y = 0.
Section 6.4
113
Taking the Laplace transform, applying the I.C. and using Theorem 6.3.1 we have (s2+s + 5/4)Y(s) = (1-e-ps/2)/s2. Thus
1-e-s/2
Y(s) =
22 s (s +s+5/4)
= e-«s/)i/5 16/25 + (16/25)s-4/25 1
^ s2 s (s + 1/2) 2+1
-ps/2
= (1-e 'lH(s), where we have used partial fractions and completed the square in the denominator of the last term. Since the numerator of the last term of H
16
can be written as —[(s+1/2) - 3/4], we see that
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